Question

select all that apply.
a. (-8, \frac{7\pi}{6})
b. (8, \frac{5\pi}{6})
c. (8, -\frac{\pi}{6})
d. (-8, \frac{13\pi}{6})

Explanation:

Step1: Recall polar coordinate transformation rules

For a polar coordinate \((r,\theta)\), it can also be represented as \((r,\theta + 2n\pi)\) or \((-r,\theta+(2n + 1)\pi)\), where \(n\in Z\).
Given the point \((8,\frac{\pi}{6})\)

Step2: Check option A

For the point \((- 8,\frac{7\pi}{6})\), here \(r=-8\), \(\theta=\frac{\pi}{6}\), using the rule \((-r,\theta+(2n + 1)\pi)\) with \(n = 0\), \(\theta+(2n + 1)\pi=\frac{\pi}{6}+\pi=\frac{7\pi}{6}\). So \((-8,\frac{7\pi}{6})\) represents the same point as \((8,\frac{\pi}{6})\)

Step3: Check option B

For the point \((8,\frac{5\pi}{6})\), using the rule \((r,\theta + 2n\pi)\), if \(\theta+2n\pi=\frac{5\pi}{6}\), then \(\frac{\pi}{6}+2n\pi=\frac{5\pi}{6}\), \(2n\pi=\frac{2\pi}{3}\), \(n=\frac{1}{3}
otin Z\). So \((8,\frac{5\pi}{6})\) does not represent the same point as \((8,\frac{\pi}{6})\)

Step4: Check option C

For the point \((8,-\frac{\pi}{6})\), using the rule \((r,\theta + 2n\pi)\), if \(\theta+2n\pi=-\frac{\pi}{6}\), then \(\frac{\pi}{6}+2n\pi=-\frac{\pi}{6}\), \(2n\pi=-\frac{\pi}{3}\), \(n=-\frac{1}{6}
otin Z\). So \((8,-\frac{\pi}{6})\) does not represent the same point as \((8,\frac{\pi}{6})\)

Step5: Check option D

For the point \((-8,\frac{13\pi}{6})\), here \(r = - 8\), using the rule \((-r,\theta+(2n + 1)\pi)\), \(\theta+(2n + 1)\pi=\frac{\pi}{6}+\pi=\frac{7\pi}{6}
eq\frac{13\pi}{6}\). Using the rule \((r,\theta + 2n\pi)\) with \(r=-8\) is not in the form. If we rewrite \(\frac{13\pi}{6}=2\pi+\frac{\pi}{6}\), for \((-8,\frac{13\pi}{6})\) using \((-r,\theta+(2n + 1)\pi)\) with \(n = 1\), \(\theta+(2n + 1)\pi=\frac{\pi}{6}+3\pi=\frac{19\pi}{6}
eq\frac{13\pi}{6}\)

Answer:

A. \((-8,\frac{7\pi}{6})\)